If kb for nx3 is 2.5×10−6, what is the poh of a 0.175 m aqueous solution of nx3? express your answer numerically

pOH = 3.18

Explanation:

The equilibrium of a weak base as NX3 is:

NX3(aq) + H2O(l) ⇄ HNX3⁺(aq) + OH-(aq)

Where the equilibrium constant, Kb, is:

Kb = 2.5x10⁻⁶ = [HNX3⁺] [OH-] / [NX3]

As both HNX3⁺ and OH- ions comes from the same equilibrium, their concentrations are the same, that is:

[HNX3⁺] = X

[OH-] = X

And: [NX3] = 0.175M:

2.5x10⁻⁶ = [X] [X] / [0.175M]

4.375x10⁻⁷ = X²

X = 6.61x10⁻⁴M = [OH-]

As pOH = -log [OH-]

yes

ya te ayud espera

explanation:

two years ago, when picking flowers in the mountains back of yosemite valley, i found a book. it was blotted and storm-beaten; all of its outer pages were mealy and crumbly. the paper seemed to dissolve like the snow beneath which it had been buried, however; many of the inner pages were well preserved. and although all were more or less stained and torn, whole chapters were easily readable. this is the condition of the great open book of yosemite glaciers today; its granite pages have been torn and blurred by the same storms that wasted the castaway book.

two years ago, when picking flowers in the mountains back of yosemite valley; i found a book. it was blotted and storm-beaten, all of its outer pages were mealy and crumbly. the paper seemed to dissolve like the snow beneath which it had been buried, however, many of the inner pages were well preserved. and although all were more or less stained and torn; whole chapters were easily readable. this is the condition of the great open book of yosemite glaciers today, its granite pages have been torn and blurred by the same storms that wasted the castaway book.